Q. The value of current in a triode valve is given by $I_{p} = 0.004\left( V_{p} + 10V_{g} \right)^{3/2}\ \ mA$ . When plate potential and grid potential are 120 V and $- 2$V respectively, then the value of mutual conductance will be
A) 600 mho
B) 60 mho
C) 6 mho
D) $6 \times 10^{- 4}\ mho$
Explanation
(d) $I_{p} = 0.004\left( V_{p} + 10V_{g} \right)^{3/2} \times 10^{- 3}$ …(i)
Differentiating it w.r.t. $V_{g}$, keeping $V_{p}$ constant, we have
$\frac{\mathrm{\Delta}I_{p}}{\mathrm{\Delta}V_{g}} = 0.004 \times \frac{3}{2}\left( V_{p} + 10\ V_{g} \right)^{1/2} \times 10 \times 10^{- 3}$
or $g_{m} = 0.004 \times \frac{3}{2}\left\lbrack 120 + 10 \times ( - 2) \right\rbrack^{1/2} \times 10^{- 2}$
$= 6 \times 10^{- 4}\ mho$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.